Law of quadratic reciprocity

We use Gauss's lemma as a stepping stone: for an odd prime $p$ and $\gcd(a,p)=1$, look at the least positive residues of $a, 2a, \dots, \frac{p-1}{2}a$ modulo $p$, and let $\mu$ be how many of them exceed $p/2$. Gauss's lemma states $\left(\dfrac{a}{p}\right)=(-1)^{\mu}$; it follows from pairing each such "large" residue $r$ with $p-r\le p/2$ and tracking signs when multiplying the $\frac{p-1}{2}$ numbers together in two different ways.

Eisenstein's refinement expresses $\mu$ as a lattice-point count: one shows $\mu \equiv \sum_{k=1}^{(p-1)/2} \left\lfloor \frac{kq}{p} \right\rfloor \pmod 2$ when $a=q$ is odd, by comparing $\lfloor kq/p\rfloor$ (the number of multiples of $p$ below $kq$) to how far $kq \bmod p$ sits from $p/2$. Applying the same counting argument symmetrically gives $\left(\frac{q}{p}\right)=(-1)^{S(q,p)}$ and $\left(\frac{p}{q}\right)=(-1)^{S(p,q)}$, where $S(q,p)=\sum_{k=1}^{(p-1)/2}\left\lfloor \frac{kq}{p}\right\rfloor$ and $S(p,q)=\sum_{k=1}^{(q-1)/2}\left\lfloor \frac{kp}{q}\right\rfloor$.

Geometrically, $S(q,p)+S(p,q)$ counts the lattice points $(x,y)$ with $1\le x\le \frac{p-1}{2}$, $1\le y\le \frac{q-1}{2}$ lying strictly below the line $qx=py$ (that count is $S(q,p)$) plus those strictly above it (that count is $S(p,q)$, by the symmetric roles of $p,q$). No lattice point lies exactly on the line since $\gcd(p,q)=1$ and $x<p$, so together these two counts exhaust the full rectangle of $\frac{p-1}{2}\cdot\frac{q-1}{2}$ lattice points.

Therefore $S(q,p)+S(p,q) = \frac{p-1}{2}\cdot\frac{q-1}{2}$, and multiplying the two Legendre-symbol formulas gives $\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{S(p,q)+S(q,p)}=(-1)^{\frac{p-1}{2}\cdot\frac{q-1}{2}}$, which is exactly $\left(\dfrac{p}{q}\right)\left(\dfrac{q}{p}\right) = (-1)^{\frac{p-1}{2}\cdot\frac{q-1}{2}}$.